# Efield integral

mathnerd15
this is the E field above a square loop with side=a at distance z on the z axis. by symmetry the Ex, Ey field cancel out. it really kind of bothers me that I can't see this substitution, including the sin(tan^-1(u)) portion. I guess it's an easy technique but I wonder how you get the substitution?

$$Ez=\frac{4\lambda z}{4\pi \varepsilon o}\int_{-a/2}^{a/2}\frac{ dx}{(z^2+x^2+a^2/4)^{3/2}}, x=\sqrt{a^2/4+z^2}tanu, dx=\sqrt{a^2/4+z^2}sec^{2}udu, I=\frac{1}{a^2/4+z^2}\int cosudu=\frac{1}{a^2/4+z^2}sinu\therefore Ez=\frac{8\lambda az}{4\pi\epsilon o \sqrt{2a^2+4z^2}z^2+a^2/4}$$

thanks!

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Thumb rule:
"To remove squares within square roots, use a suitable trig substitution. If stuff is more difficult than that, forget about it"

mathnerd15
I try to do these by hand- I wonder if some people do all of these by hand?

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Homework Helper
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I wonder if some people do all of these by hand?

Those who need to learn them, such as students. Those who are professionally competent don't, but at most, picks up a table of integrals. Only those are professionally competent who has learned them. mathnerd15
I'm not sure if maybe it's better to do a lot of problems out of mathematical and Schaum books. I had another career before this that was completely different than physics/mathematics
I've been looking at Apostol and Hubbard Calculus (introduces manifolds) and I'm not sure if I should study these since I've already done Stewart (problems are easy I know)

Homework Helper
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I'm not sure what you're getting at.

If lots of the basic tools used within research seems utterly magical to you, because you haven't learnt the logic behind them, can you ever succeed as a researcher? There's a reason why an education within physics/math is a fast-track over the evolution of the subjects, ordered in a pedagogically optimal way.
Alternatively: Why should anyone want to employ you, unless you have a proven ability to understand and master previous areas of research?

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Thus, the function behind drilling students with such as nasty integrals is at least three-fold:
To hone the mind of the student, and familiarize him with the technical language, along with being a quality control of the student.

mathnerd15
thanks, it's not a difficult substitution.

so the integral substitution works well with the tan^2+1 term reducing it to cosu and you just substitute back in for u and solve,
$$sin(arctan(\frac{x}{(a^2/4+z^2)^{1/2}}))=\frac{x}{(a^2/4+x^2+z^2)^{1/2}}$$

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